If ln (x + y) = 2xy,then y'(0) = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the equation $\ln (x + y) = 2xy$. First,find the value of $y$ when $x = 0$: $\ln (0 + y) = 2(0)y \implies \ln (y) = 0 \implies y = e^0 = 1$.

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the equation $\ln (x + y) = 2xy$. First,find the value of $y$ when $x = 0$: $\ln (0 + y) = 2(0)y \implies \ln (y) = 0 \implies y = e^0 = 1$. Now,differentiate both sides with respect to $x$: $\frac{d}{dx} [\ln (x + y)] = \frac{d}{dx} [2xy]$ $\frac{1}{x + y} \cdot (1 + y') = 2(y + xy')$ Substitute $x = 0$ and $y = 1$ into the differentiated equation: $\frac{1}{0 + 1} \cdot (1 + y'(0)) = 2(1 + 0 \cdot y'(0))$ $1 \cdot (1 + y'(0)) = 2(1)$ $1 + y'(0) = 2$ $y'(0) = 2 - 1 = 1$.

If $\ln (x + y) = 2xy$,then $y'(0) =$

Similar Questions

If $3\sin (xy) + 4\cos (xy) = 5$,then $\frac{dy}{dx} = $

If $xy = \tan^{-1}(xy) + \cot^{-1}(xy)$,then $\left(\frac{dy}{dx}\right)_{(4,2)} = ?$ (where $x, y \in \mathbb{R}$)

Find $\frac{dy}{dx}$,if $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$.

Find $\frac{dy}{dx}$ if $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$.

If ${x^y} \cdot {y^x} = 1$,then ${{dy} \over {dx}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module